# How do you determine the oxidizing or reducing agent in a redox equation?

Feb 29, 2016

The species whose oxidation number is increased is the reducing agent, and the species whose oxidation number is decreased in the oxidizing agent.

#### Explanation:

Let's take a simple example of combustion, which is formally a redox reaction:

$C \left(s\right) + {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right)$

Carbon is oxidized from elemental carbon, the zerovalent, elemental state (oxidation state = $0$), to $C \left(I V\right)$, its maximum oxidation state. Likewise, oxygen has been REDUCED from the zerovalent state, $O$ to $- I I$.

Since carbon has LOST electrons (formally) it has been oxidized by definition, and the species which ACCEPTED those electrons was (i) the oxidizing agent, and (ii) has been reduced.

Redox reactions formalize the addition and subtractions of electrons by the writing of half equations, in which separate oxidation and reduction by electron transfer is explicit.

I could represent the above reaction in these terms,

$C \left(s\right) \rightarrow {C}^{4 +} + 4 {e}^{-}$; $\text{C is Oxidized}$, and has therefore donated electrons formally!

${O}_{2} \left(g\right) + 4 {e}^{-} \rightarrow 2 {O}^{2 -}$; $\text{O is Reduced}$, and has therefore accepted electrons formally!

The sum of the redox half equation represent the overall reaction presented above.

So the moral?

$\text{Loss of electrons = OXIDATION}$

$\text{Gain in electrons = REDUCTION}$

To reiterate if something has lost electrons, it is the reducing agent (because the electrons have to go somewhere!). And if something has gained electrons, it is the oxidizing agent. Oxygen gas is thus a superb oxidizing agent.